Abstract
The classical skew-product decomposition of planar Brownian motionrepresents the process in polar coordinates as an autonomously Markovian part and an part that is an independent Brownian motion on the unit circle time-changed according to the part. Theorem 4 of L09 gives a broad generalization of this fact to a setting where there is a diffusion on a manifold $X$ with a distribution that is equivariant under the smooth action ofa Lie group $K$. Under appropriate conditions, there is a decomposition into an autonomously Markovian radial part that lives on the space of orbits of $K$ and an angular part that is an independent Brownian motion on the homogeneous space $K/M$, where $M$ is the isotropy subgroup of a point of $x$, that is time-changed with a time-change that is adapted to the filtration of the part. We present two apparent counterexamples to Theorem 4 of L09. In the first counterexample the part is not a time-change of any Brownian motionon $K/M$, whereas in the second counterexample the part is the time-change of a Brownian motion on $K/M$ but this Brownian motion is not independent of the part. In both of these examples $K/M$ has dimension $1$. The statement and proof of Theorem 4 in L09 remain valid when $K/M$ has dimension greater than $1$. Our examples raise the question of what conditions lead to the usual sort of skew-product decomposition when $K/M$ has dimension $1$ and what conditions lead to there being no decomposition at all or one in which the part is a time-changed Brownian motion but this Brownian motion is not independent of the part.
Highlights
The archetypal skew-product decomposition of a Markov process is the decomposition of a Brownian motion in the plane (Bt)t≥0 into its radial and angular part (1.1)Bt = |Bt| exp(iθt).Here the radial part (|Bt|)t≥0 is a two-dimensional Bessel process and θt = yτt, wheret≥0 is a one-dimensional Brownian motion that is independent of the radial part (|Bt|)t≥0 and τ is a time-change that is adapted to the filtration generated by the process|B|
We found an apparent counterexample to the main result, Theorem 4 of that paper in which there is a decomposition of the process into an autonomously Markov radial process on Y and an angular part that is a Brownian motion on K/M time-changed according to the radial process, but this Brownian motion is not, contrary to the claim of [Lia09], independent of the radial process, see Section 4 for an exposition of the counterexample
We show that in this case that there is no skew-product decomposition for a somewhat different reason: the angular part oft≥0 cannot be written as a time-changed Brownian motion on the unit circle in the plane
Summary
The archetypal skew-product decomposition of a Markov process is the decomposition of a Brownian motion in the plane (Bt)t≥0 into its radial and angular part (1.1). We found an apparent counterexample to the main result, Theorem 4 of that paper in which there is a decomposition of the process into an autonomously Markov radial process on Y and an angular part that is a Brownian motion on K/M time-changed according to the radial process, but this Brownian motion is not, contrary to the claim of [Lia09], independent of the radial process, see Section 4 for an exposition of the counterexample This seeming contradiction appears because the assumption from [Lia09] that K/M is irreducible is not strong enough to ensure the nonexistence of a nonzero M -invariant tangent vector in the special case when, as in our construction, K/M has dimension 1. A sampling of other results involving skew-products can be found in [Tay, LCO09, ELJL10, BN06]
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